epiR (version 0.9-82)

epi.equivc: Estimate the sample size for a parallel equivalence trial, continuous outcomes

Description

Computes the sample size for a parallel equivalence trial with a continuous outcome variable.

Usage

epi.equivc(treat, control, sd, delta, n, r = 1, power, alpha)

Arguments

treat

the expected mean of the outcome of interest in the treatment group.

control

the expected mean of the outcome of interest in the control group.

sd

the expected population standard deviation of the outcome of interest.

delta

the equivalence limit, expressed as a proportion.

n

scalar, the total number of study subjects in the trial.

r

scalar, the number in the treatment group divided by the number in the control group.

power

scalar, the required study power.

alpha

scalar, defining the desired alpha level.

Value

A list containing one or more of the following:

n.treat

the required number of study subject in the treatment group.

n.control

the required number of study subject in the control group.

n.total

the total number of study subjects required.

References

Bennett JE, Dismukes WE, Duma RJ, Medoff G, Sande MA, Gallis H, Leonard J, Fields BT, Bradshaw M, Haywood H, McGee Z, Cate TR, Cobbs CG, Warner JF and Alling DW (1979). A comparison of amphotericin B alone and combined flucytosine in the treatment of cryptococcal meningitis. New England Journal of Medicine 301: 126 - 131.

Chow S, Shao J, Wang H (2008). Sample Size Calculations in Clinical Research. Chapman & Hall/CRC Biostatistics Series, pp. 91.

Ewald B (2013). Making sense of equivalence and non-inferiority trials. Australian Prescriber 36: 170 - 173.

Julious SA (2004). Sample sizes for clinical trials with normal data. Statistics in Medicine 23: 1921 - 1986.

Julious SA (2009). Estimating Samples Sizes in Clinical Trials. CRC, New York.

Machin D, Campbell MJ, Tan SB, Tan SH (2009). Sample Size Tables for Clinical Studies. Wiley Blackwell, New York.

Examples

## EXAMPLE 1 (from Machin, Campbell, Tan and Tan 2009 p. 113):
## It is anticipated that patients on a particular drug have a mean diastolic 
## blood pressure of 96 mmHg, as against 94 mmHg on an alternative. It is also 
## anticipated that the standard deviation of diastolic BP is approximately 
## 8 mmHg. If one wishes to confirm that the difference is likely to be less 
## than 5 mmHg, that is, one wishes to show equivalence, how many patients 
## are need to be enrolled in the trial? Assume 80% power and 
## 95% significance.

epi.equivc(treat = 94, control = 96, sd = 8, delta = 5, n = NA, 
   r = 1, power = 0.80, alpha = 0.05)

## A total of 244 subjects need to be enrolled in the trial, 122 in the 
## treatment group and 122 in the control group.

## EXAMPLE 2 (from Chow S, Shao J, Wang H 2008, p. 64):
## A pharmaceutical company is interested in conducting a clinical trial
## to compare two cholesterol lowering agents for treatment of patients with
## congestive heart disease using a parallel design. The primary efficacy 
## parameter is the LDL. In what follows, we will consider the situation
## where the intended trial is for testing equivalence of mean responses 
## in LDL. Assume that 80% power is required at a 5% level of significance.

## In this example, we assume a 5 unit (i.e. delta = 5) change of LDL is 
## considered of clinically meaningful difference. Assume the standard 
## of LDL is 10 units and the LDL concentration in the treatment group is 20 
## units and the LDL concentration in the control group is 21 units.

epi.equivc(treat = 20, control = 21, sd = 10, delta = 5, n = NA, 
   r = 1, power = 0.80, alpha = 0.05)

## A total of 216 subjects need to be enrolled in the trial, 108 in the 
## treatment group and 108 in the control group.

## EXAMPLE 2 (cont.):
## Suppose only 150 subjects were enrolled in the trial, 75 in the treatment
## group and 75 in the control group. What is the estimated study power?

epi.equivc(treat = 0.20, control = 0.21, sd = 0.10, delta = 0.05, n = 150, 
   r = 1, power = NA, alpha = 0.05)

## With only 150 subjects the estimated study power is 0.58.